JellyBean98
Member
how do I calculate how to step 12volts down to 4.5volts? or does anyone know what resistors I will need? thanks
please help!! changing 12v to 4.5v
- Thread starter JellyBean98
- Start date
it all depends on how many amps the device will suck down. the formula is voltage = current x resistance. so you know the voltage is constant (basically), and if you know the current, just divide the voltage by the current to get the resistance
but keep in mind that wattage = current x voltage, so you have to take that into consideration when buying a resistor, to make sure it's a high enough wattage
but keep in mind that wattage = current x voltage, so you have to take that into consideration when buying a resistor, to make sure it's a high enough wattagejohnsharpe
Member
- :
- 2003 P5
It's been a while, but I think I got this right...
The resistor divider eqn is simple:
V1 = (R1 X E) / R2
in this setup:
E ------\/\/\/--------> E'
R1 |
|
>
R2 >
>
|
//// gnd
E is your incoming voltage (12V in this case)
E' is the desired output (4.5V in this case)
edit: Hmm...drawing didn't quite come out like I wanted...R2 should go to ground between R1 and E'...sorry 'bout that. I'll try to find a decent pic...
By manipulating the equation, you can figure out whatever value you need.
HOWEVER!! As got wake mentioned, all depends on how much current the device will suck....you need to take into account the power dissipation limits of the resistors, otherwise you'll let the smoke out of them.
For this application, I agree with Pheonix...spend the extra $1 and buy a proper regulator. LM7805 will output 5V and can handle upto 800mA without getting really hot. Make sure however that you've got a good heat sink. Dropping 7V will make it warm if you're running upto that load (and prolong it's life).
For higher loads or if you want exactly 4.5V, go to an adjustable reg, LM350T is adjustable from 1.25V to Vcc-1.25V by using external resistors and can handle upto 3A.
The most important thing to do is test it somewhere other than the final intended location...that way you only risk loosing some cheap parts and not an entire stereo/car/etc...
Good luck, let us know if you need any more info!
John
The resistor divider eqn is simple:
V1 = (R1 X E) / R2
in this setup:
E ------\/\/\/--------> E'
R1 |
|
>
R2 >
>
|
//// gnd
E is your incoming voltage (12V in this case)
E' is the desired output (4.5V in this case)
edit: Hmm...drawing didn't quite come out like I wanted...R2 should go to ground between R1 and E'...sorry 'bout that. I'll try to find a decent pic...
By manipulating the equation, you can figure out whatever value you need.
HOWEVER!! As got wake mentioned, all depends on how much current the device will suck....you need to take into account the power dissipation limits of the resistors, otherwise you'll let the smoke out of them.
For this application, I agree with Pheonix...spend the extra $1 and buy a proper regulator. LM7805 will output 5V and can handle upto 800mA without getting really hot. Make sure however that you've got a good heat sink. Dropping 7V will make it warm if you're running upto that load (and prolong it's life).
For higher loads or if you want exactly 4.5V, go to an adjustable reg, LM350T is adjustable from 1.25V to Vcc-1.25V by using external resistors and can handle upto 3A.
The most important thing to do is test it somewhere other than the final intended location...that way you only risk loosing some cheap parts and not an entire stereo/car/etc...
Good luck, let us know if you need any more info!
John
Last edited:
traitorhound
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i second the use of a voltage regulator. ...and they put off less heat then resistors
